Check the picture below.
[tex]sin(EDG )=\cfrac{\stackrel{opposite}{10}}{\underset{hypotenuse}{10\sqrt{3}}}\implies sin(EDG )=\cfrac{1}{\sqrt{3}}\implies sin(EDG )=\cfrac{1}{\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}} \\\\\\ \stackrel{\textit{rationalizing the denominator}}{sin(EDG )=\cfrac{\sqrt{3}}{\sqrt{3^2}}\implies sin(EDG )=\cfrac{\sqrt{3}}{3}} \\\\[-0.35em] ~\dotfill[/tex]
[tex]cos(EDG )=\cfrac{\stackrel{adjacent}{10\sqrt{2}}}{\underset{hypotenuse}{10\sqrt{3}}}\implies cos(EDG )=\cfrac{\sqrt{2}}{\sqrt{3}}\implies cos(EDG )=\cfrac{\sqrt{2}}{\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}} \\\\\\ \stackrel{\textit{rationalizing the denominator}}{cos(EDG )=\cfrac{\sqrt{6}}{\sqrt{3^2}}\implies cos(EDG )=\cfrac{\sqrt{6}}{3}}[/tex]
